Question

find the area of the region enclosed by the curves $y^2 - 4x = 4$ and $x - y = 2$. the area of the region enclosed by the curves is $dfrac{64}{3}$ (simplify your answer.)

Explanation:

Step1: Express x from both equations

From \( y^2 - 4x = 4 \), we get \( x=\frac{y^2 - 4}{4} \). From \( x - y = 2 \), we get \( x = y + 2 \).

Step2: Find intersection points

Set \( \frac{y^2 - 4}{4}=y + 2 \). Multiply both sides by 4: \( y^2 - 4 = 4y + 8 \). Rearrange to \( y^2 - 4y - 12 = 0 \). Factor: \( (y - 6)(y + 2)=0 \). So \( y = 6 \) or \( y=-2 \).

Step3: Set up the integral for area

The area \( A=\int_{-2}^{6}[(y + 2)-\frac{y^2 - 4}{4}]dy \). Simplify the integrand: \( y + 2-\frac{y^2}{4}+1=-\frac{y^2}{4}+y + 3 \).

Step4: Integrate

Integrate term - by - term: \( \int_{-2}^{6}-\frac{y^2}{4}dy+\int_{-2}^{6}y\ dy+\int_{-2}^{6}3\ dy \).

• \( \int-\frac{y^2}{4}dy=-\frac{y^3}{12}+C \), evaluated from - 2 to 6: \( [-\frac{6^3}{12}+\frac{(-2)^3}{12}]=[ - 18-\frac{2}{3}]=-\frac{56}{3} \)
• \( \int y\ dy=\frac{y^2}{2}+C \), evaluated from - 2 to 6: \( [\frac{6^2}{2}-\frac{(-2)^2}{2}]=[18 - 2]=16 \)
• \( \int 3\ dy = 3y+C \), evaluated from - 2 to 6: \( [3\times6-3\times(-2)]=[18 + 6]=24 \)

Step5: Sum the integrals

\( A=-\frac{56}{3}+16 + 24 \). Convert 16 and 24 to thirds: \( 16=\frac{48}{3} \), \( 24=\frac{72}{3} \). Then \( A=-\frac{56}{3}+\frac{48}{3}+\frac{72}{3}=\frac{-56 + 48+72}{3}=\frac{64}{3} \).

Answer:

\(\frac{64}{3}\)